Asked by Jenna
A car is traveling at 70 mph skids 240 ft after it's breaks are suddenly applied. Assuming that the cars breaking system provides a constant deceleration, find that deceleration. How long did it take the car to stop?
Answers
Answered by
Damon
You want me to do this with calculus with these terrible units evidently so I will.
70 mph (1h/3600s)(5280ft/m) = 103 ft/s
d^2x/dt^2 = a
dx/dt = a t + Vi where Vi is initial speed
x = (1/2) a t^2 + Vi t + Xo wher Xo is initial x (call it 0 here
so
dx/dt at end = 0
0 = a t + 70 so a = -103/t
then
240 ft = (1/2) at^2 + 103 t
240 = (1/2)(-103/t) t^2 + 103 t
240 = -51.5 t + 103 t = 51.5 t
t = 4.66 seconds to stop
a = -103/4.66 = -22.1 ft/s^2
240 =
70 mph (1h/3600s)(5280ft/m) = 103 ft/s
d^2x/dt^2 = a
dx/dt = a t + Vi where Vi is initial speed
x = (1/2) a t^2 + Vi t + Xo wher Xo is initial x (call it 0 here
so
dx/dt at end = 0
0 = a t + 70 so a = -103/t
then
240 ft = (1/2) at^2 + 103 t
240 = (1/2)(-103/t) t^2 + 103 t
240 = -51.5 t + 103 t = 51.5 t
t = 4.66 seconds to stop
a = -103/4.66 = -22.1 ft/s^2
240 =